Empirical formula 12.67 aluminum, 19.73 nitrogen, 67.60 oxygen
Do you mean these are percents?
Take a 100 grams sample which will give you
12.67 g Al
19.73 g N
67.6 g O
Convert each to mol (mol = g/atomic mass), then find the ratio of the elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself. Round to whole numbers but don't round too much; i.e., 3.5 rounds to 7 (2*3.5).
To find the empirical formula, you need to determine the simplest whole number ratio of the elements in the compound. Here's the step-by-step process to determine the empirical formula using the given percentage composition:
Step 1: Convert the percentage of each element into grams. To do this, assume you have a 100g sample of the compound.
- Aluminum: (12.67% / 100%) * 100g = 12.67g
- Nitrogen: (19.73% / 100%) * 100g = 19.73g
- Oxygen: (67.60% / 100%) * 100g = 67.60g
Step 2: Determine the moles of each element using their molar masses.
- Aluminum (Al): Molar mass = 26.98 g/mol; moles = 12.67g / 26.98 g/mol = 0.469 moles
- Nitrogen (N): Molar mass = 14.01 g/mol; moles = 19.73g / 14.01 g/mol = 1.41 moles
- Oxygen (O): Molar mass = 16.00 g/mol; moles = 67.60g / 16.00 g/mol = 4.23 moles
Step 3: Find the smallest whole number ratio by dividing each mole value by the smallest moles value. In this case, the smallest moles value is 0.469 moles.
- Aluminum: 0.469 moles / 0.469 moles = 1
- Nitrogen: 1.41 moles / 0.469 moles ≈ 3
- Oxygen: 4.23 moles / 0.469 moles ≈ 9
Step 4: Write the empirical formula using the whole number ratios obtained in step 3.
The empirical formula is AlN3O9
To find the empirical formula, you need to determine the ratio of the elements in the compound based on their mass percentages. In this case, we are given the following percentages:
Aluminum: 12.67%
Nitrogen: 19.73%
Oxygen: 67.60%
Step 1: Convert the percentages to grams.
To make calculations easier, we assume that we have 100 grams of the compound. Therefore, we have:
Aluminum: 12.67 grams
Nitrogen: 19.73 grams
Oxygen: 67.60 grams
Step 2: Convert the grams to moles.
To convert grams to moles, divide the given mass by the molar mass of each element.
The molar mass of aluminum (Al) is 26.98 g/mol.
The molar mass of nitrogen (N) is 14.01 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.
Calculations:
Aluminum: 12.67 g / 26.98 g/mol = 0.469 mol
Nitrogen: 19.73 g / 14.01 g/mol = 1.407 mol
Oxygen: 67.60 g / 16.00 g/mol = 4.225 mol
Step 3: Determine the mole ratio.
Divide each number of moles by the smallest number of moles to find the simplest, whole number ratio.
Calculations:
Aluminum: 0.469 mol / 0.469 mol = 1
Nitrogen: 1.407 mol / 0.469 mol = 3
Oxygen: 4.225 mol / 0.469 mol = 9
The empirical formula is AlN₃O₉, which represents the simplest, whole number ratio of aluminum, nitrogen, and oxygen in the compound.